Problem: $h(n) = -4n-2$ $g(t) = 4t^{3}-3t^{2}+3(h(t))$ $f(t) = 6t^{2}+3(h(t))$ $ h(f(-1)) = {?} $
Explanation: First, let's solve for the value of the inner function, $f(-1)$ . Then we'll know what to plug into the outer function. $f(-1) = 6(-1)^{2}+3(h(-1))$ To solve for the value of $f$ , we need to solve for the value of $h(-1)$ $h(-1) = (-4)(-1)-2$ $h(-1) = 2$ That means $f(-1) = 6(-1)^{2}+(3)(2)$ $f(-1) = 12$ Now we know that $f(-1) = 12$ . Let's solve for $h(f(-1))$ , which is $h(12)$ $h(12) = (-4)(12)-2$ $h(12) = -50$